Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 10 - Rotation - Problems: 12b

Answer

$420\ rev$

Work Step by Step

Initial angular acceleration speed $\omega_i=1200\ rev/s$ Final angular acceleration speed $\omega_f=3000\ rev/s$ Angular acceleration $\alpha =9000\ rev/min^2$ From rotational kinematics equation, we have: $\omega_f^2-\omega_i^2 = 2\alpha\theta$ $(3000\ rev/min)^2 - (1200\ rev/min)^2 = 2 (9000\ rev/min^2)\theta$ $7560000 (rev/min)^2 = 2 (9000\ rev/min^2)\theta$ $\theta = 420\ rev$
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